v=i^+2j^
⇒x=t ..... (i)
y=2t−21(10t2) ..... (ii)
From (i) and (ii)
y=2x−5x2
A projectile is given an initial velocity of (i^+2j^)ms−1, where i^ is along the ground and j^ is along the vertical upward. If g=10ms−2, the equation of its trajectory is :
Held on 7 Apr 2013 · Verified 6 Jul 2026.
4y=2x−5x2
4y=2x−25x2
y=x−5x2
y=2x−5x2
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